## Elementary Technical Mathematics

Published by Brooks Cole

# Chapter 16 - Cumulative Review - Page 568: 2

#### Answer

The number to the nearest three significant digits is $37.0\times {{10}^{-9}}$.

#### Work Step by Step

$\frac{\left( 612\times {{10}^{-6}} \right)\left( 15\times {{10}^{-9}} \right)\left( 2.7\times {{10}^{3}} \right)}{\left( 82\times {{10}^{9}} \right)\left( 8.16\times {{10}^{-12}} \right)}$, Combine the powers of $10$ by using the formula $\frac{{{a}^{x}}\cdot {{a}^{y}}}{{{a}^{z}}}={{a}^{x+y-z}}$. \begin{align} & \frac{\left( 612\times {{10}^{-6}} \right)\left( 15\times {{10}^{-9}} \right)\left( 2.7\times {{10}^{3}} \right)}{\left( 82\times {{10}^{9}} \right)\left( 8.16\times {{10}^{-12}} \right)}=\frac{612\times 15\times 2.7\times {{10}^{-6-9+3}}}{82\times 8.16\times {{10}^{9-12}}} \\ & =\frac{612\times 15\times 2.7\times {{10}^{-12}}}{82\times 8.16\times {{10}^{-3}}} \\ & =\frac{612\times 15\times 2.7\times {{10}^{-12+3}}}{82\times 8.16} \end{align} Further simplify the above expression. \begin{align} & \frac{\left( 612\times {{10}^{-6}} \right)\left( 15\times {{10}^{-9}} \right)\left( 2.7\times {{10}^{3}} \right)}{\left( 82\times {{10}^{9}} \right)\left( 8.16\times {{10}^{-12}} \right)}=\frac{612\times 15\times 2.7\times {{10}^{-12+3}}}{82\times 8.16} \\ & =37.04\times {{10}^{-9}} \end{align}

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